Question: $ g(x) = \int_{1}^{x}(3t^2 +4t )\,dt\,$ $ g\,'(2)\, =$
The Fundamental Theorem of Calculus If $~ g(x)=\int_a^xf(t)\,dt\,$, then $~g^\prime (x)=f(x)\,$ This only works if $f$ is continuous on $[a,b]$. Thankfully, the function $f(t) = 3t^2 + 4t$ is continuous on $[1,2]$. Applying the theorem We're given: $ g(x) = \int_{1}^{x}(3t^2 +4t )\,dt\,$ So the theorem tells us: $ g\,^\prime(x) =3x^2 +4x $ Evaluating $g'(2)$ $ g'(2)= 3(2)^2 + 4(2) = 12 + 8 = 20$ The answer: $g'(2)=20$